# How do you integrate (x^2-1)/(x^4-16) using partial fractions?

Sep 21, 2016

$\frac{3}{32} \ln | \frac{x - 2}{x + 2} | + \frac{5}{16} a r c \tan \left(\frac{x}{2}\right) + C$.

#### Explanation:

Let $I = \int \frac{{x}^{2} - 1}{{x}^{4} - 16} \mathrm{dx}$.

Here, the Integrand $\frac{{x}^{2} - 1}{{x}^{4} - 16} = \frac{{x}^{2} - 1}{\left({x}^{2} - 4\right) \left({x}^{2} + 4\right)}$

$= \frac{y - 1}{\left(y - 4\right) \left(y + 4\right)} , w h e r e , y = {x}^{2}$ We split the Integrand using the

Partial Fractions. So, let,

$\frac{y - 1}{\left(y - 4\right) \left(y + 4\right)} = \frac{A}{y - 4} + \frac{B}{y + 4} , w h e r e , A , B \in \mathbb{R}$.

To find, A, &, B, we use Heavyside's Cover-up Method :-

$A = {\left[\frac{y - 1}{y + 4}\right]}_{y = 4} = \frac{4 - 1}{4 + 4} = \frac{3}{8}$.

$B = {\left[\frac{y - 1}{y - 4}\right]}_{y = - 4} = \frac{- 4 - 1}{- 4 - 4} = \frac{5}{8}$. Hence,

$\frac{y - 1}{\left(y - 4\right) \left(y + 4\right)} = \frac{\frac{3}{8}}{y - 4} + \frac{\frac{5}{8}}{y + 4}$. Letting, $y = {x}^{2}$,

$\frac{{x}^{2} - 1}{{x}^{4} - 16} = \frac{\frac{3}{8}}{{x}^{2} - 4} + \frac{\frac{5}{8}}{{x}^{2} + 4} .$ Therefore,

$I = \frac{3}{8} \int \frac{1}{{x}^{2} - {2}^{2}} \mathrm{dx} + \frac{5}{8} \int \frac{1}{{x}^{2} + {2}^{2}} \mathrm{dx}$

$= \frac{3}{8} \cdot \frac{1}{2 \cdot 2} \ln | \frac{x - 2}{x + 2} | + \frac{5}{8} \cdot \frac{1}{2} a r c \tan \left(\frac{x}{2}\right)$

$\therefore I = \frac{3}{32} \ln | \frac{x - 2}{x + 2} | + \frac{5}{16} a r c \tan \left(\frac{x}{2}\right) + C$.

Enjoy Maths.!